ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)

\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)

https://sg.docworkspace.com/l/sIM-LioBEocfloAY

a)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)

\(=x^3+2x^2-9x-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)

\(=-5x^3-38x^2-76x-46\)

b)\(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)

\(=2x^3-16x^2+32x-\left(x^3+5x^2-4x-20\right)+2x^2+20x+50-x^2+2x-1\)

\(=x^3-20x^2+58x+69\)

c)\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-\left(2x^3-x^2-18x+9\right)\)

\(=-18x^3-46x^2-8x+16\).

a) Ta có: \(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)

\(=x^3+2x^2-9x-18-\left(x^3-3x-x^2+3\right)-5x^3-40x^2-80x-x^2+10x-25\)

\(=-4x^3-39x^2-79x-43-x^3+3x+x^2-3\)

\(=-5x^3-38x^2-76x-46\)

b) Ta có: \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2x^2+20x+50-x^2+2x-1\)

\(=2x^3-16x^2+32x-x^3+4x-5x^2+20+x^2+22x+49\)

\(=x^3-20x^2+56x+49\)

c) Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x-3\right)\left(x+3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3+48x-36x-2x^3+18x+x^2-9\)

\(=-18x^3+2x^2+40x+16\)

a, Ta có : \(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x-5\right)^2\)

\(=x^3-9x+2x^2-18-\left(x^3-x^2-3x+3\right)-\left(5x^3+40x^2+80x\right)-\left(x^2-10x+25\right)\)

\(=x^3-9x+2x^2-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)

\(=-5x^3-38x^2-76x-46\)

\(b,\) Ta có : \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x-4\right)^2-\left(x+5\right)\left(x^2-4\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x^3+5x^2-4x-20\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)

\(=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2+20x+50-x^2+2x-1\)

\(=x^3-20x^2+58x+69\)

c, Ta có :\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x-3\right)\left(x+3\right)\)

\(=\left(x+5\right)^2-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=\left(x+5\right)^2-\left(16x^3+48x^2+36x\right)-\left(2x^3-x^2-18x+9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+x^2+18x-9\)

\(=-18x^3-46x^2-8x+16\)

       x²-4x+4=4x²-12x+9

\(\Leftrightarrow\)3x²-8x+5=0

\(\Leftrightarrow\)3x²-3x-5x+5=0

\(\Leftrightarrow\)3x(x-1)-5(x-1)=0

\(\Leftrightarrow\)(x-1)(3x-5)=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{3}\\x=1\end{cases}}\)

b,x²-2x-25=0

\(\Leftrightarrow\)(x-1)²-26=0

\(\Leftrightarrow\)(x-1-\(\sqrt{26}\))(x-1+\(\sqrt{26}\))=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{26}+1\\x=-\sqrt{26}+1\end{cases}}\)

2, a, x^2-2x+1+4=(x-1)^2+4\(\ge\)4

b, 4x^2-4x+1-1+y^2+2y+1-1-2015=(2x-1)^2+(y+1)^2-2017\(\ge\)-2017

mk làm như thế thôi chứ bài kia dài quá mk làm biếng sory

Nguyễn Thị Hà Tiên : Cảm ơn bạn nhiều lắm =)) Mik đã bt hướng làm bài rồi :3 Thực sự cảm ơn pạn nek <3 

Bài 1: 

a)  \(\left(x-2\right)^2=4x^2-12x+9\Leftrightarrow\left(x-2\right)^2=\left(2x-9\right)^2\Leftrightarrow\left(x-2\right)^2-\left(2x-9\right)^2=0\)

\(\Leftrightarrow\left(x-2+2x-9\right)\left(x-2-2x+9\right)=0\Leftrightarrow\left(3x-11\right)\left(7-x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}3x-11=0\Leftrightarrow3x=11\Leftrightarrow x=\frac{11}{3}\\7-x=0\Leftrightarrow-x=-7\Leftrightarrow x=7\end{cases}}\)

VẬy tập nghiệm của phương trình là : S={11/3 ; 7}

b)   Nếu x^2 -2x  =25 thì lẻ lắm . Tớ nghĩ phải là :  x^2 -2x  = 24 

Bài 2 : 

a)  \(A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)

vì \(\left(x-1\right)^2\ge0\) nên \(\left(x-1\right)^2+4\ge4\)  hay \(A\ge4\)

Vậy GTNN của A là 4  khi x = 1        ( hay x-1 =0 )

b)  \(B=4x^2-4x+y^2+2y-2015=\left(4x^2-4x+1\right)+\left(y^2+2y+1\right)-2017\)

\(=\left(2x-1\right)^2+\left(y+1\right)^2-2017\)

Vì \(\left(2x-1\right)^2\ge0\)     và \(\left(y+1\right)^2\ge0\)   nên   \(\left(2x-1\right)^2+\left(y+1\right)^2-2017\ge-2017\)

HAy \(B\ge-2017\)    Vậy GTNN của B là -2017  khi x=1/2   và y =  -1

bài 1: 

2(x^2-9).4(x^2-1)

=(2x^2-18)(4x^2-4)

=8x^4-8x^2-72x^2+72

=8x^4-80x^2+72

\(Bai1:2\left(x-3\right)\left(x+3\right)+4\left(x-1\right)\left(x+1\right)\)

\(=2\left(x^2-9\right)+4\left(x^2-1\right)\)

\(=2x^2-18+4x^2-4\)

\(=6x^2-22\)

\(Bai2:-\left(6x-1\right)\left(3-2x\right)+\left(3x-2\right)\left(4x-3\right)=17\)

\(\Leftrightarrow-\left(18x-12x^2-3+2x\right)+12x^2-9x-8x+6=17\)

\(\Leftrightarrow-18x+12x^2+3-2x+12x^2-9x-8x+6=17\)

\(\Leftrightarrow24x^2-37x+9-17=0\)

\(\Leftrightarrow24x^2-37x-8=0\)

Đề sai??